$\Delta u = 0$ in $B^{+}\cap \{x_{n} > 0\}$ and $u = u_{x_{n}} = 0$ in $\{x_{n} = 0\}$ $\implies $ $u = 0$.

Question

Let $B^{+} = \{x \in \mathbb{R}^n ~ \colon ~ ||x|| \leq 1 ~~ \text{and} ~~ x_n >0 \}$ and $u \in C^2(B^{+})\cap C^1(\overline B^{+})$ such that

$$ \Delta u = 0 ~~ \text{ in } ~~ B^+ \hspace{3.5cm} \\ u = u_{x_n} = 0 ~~ \text{ in } ~~ \partial B^+\cap \{x_n=0\}.$$

Is it true that $ u \equiv 0$? If we had $u \geq 0$ we could use Hopf Lemma and conclude that it is true. Can we conclude that removing this hypothesis?

Answer 1

Since, $u \in C^1(\overline{B^+})$ and $u = u_{x_n} = 0$ on $\Gamma := \partial B^+ \cap \{x_n = 0\}$ then $$v := \begin{cases} u & \text{ in } B^+ \cup \Gamma \\ 0 & \text{ in } B^-\end{cases}$$ defines a $C^1$ extension of $u$ in $B^-$. It suffices to show that $v$ is harmonic in $B$ and hence by identity principle $v \equiv 0$ in $B$ proving the claim.

If we are able to show that $v$ satisfies a 'local' version of the mean-value property in the sense that for every $x \in B$ there exists $r_x > 0$ s.t., $$\frac{1}{|\partial B_r|}\int_{\partial B_r(x)} v(y)\,d\sigma(y) = v(x), \, \forall \, r \in (0,r_x) \tag {1}$$ then this shows $v$ is harmonic in neighborhood of every point and hence harmonic in $B$.

In order to show $(1)$ we note that $$\frac{d}{dr}\left(\frac{1}{|\partial B_r|}\int_{\partial B_r(x)} v(y)\,d\sigma(y)\right) = \frac{1}{|\partial B_r|}\int_{\partial B_r(x)} \frac{\partial v}{\partial n}\,d\sigma \tag {2}$$ for any $C^1$ function $v$. Thus, if we show that for every $x \in B$ there is a choice of $r_x > 0$ such that $$\int_{\partial B_r(x)} \frac{\partial v}{\partial n}\,d\sigma = 0 \tag{3}$$ for all $r \in (0,r_x)$ with $B_{r}(x) \subset B$ then we are done.

Now, if $x \in B^+$ (or $B^-$) then $v$ is harmonic in the classical sense and choosing $r_x > 0$ s.t., $B_{r_x} \subset B^+$ (or, $B^-$) for any $r \in (0,r_x)$ we have $$0 = \int_{B_r(x)} \Delta v(y) \,dy = \int_{\partial B_r(x)} \frac{\partial v}{\partial n}\,d\sigma(y)$$ from integration by parts. Now, suppose $x \in \Gamma$ then for $r_x > 0$ s.t., $B_{r_x}(x) \subset B$ we have \begin{align*} \int_{\partial B_r(x)} \frac{\partial v}{\partial n}\,d\sigma(y) &= \int_{\partial B_r(x) \cap B^+} \frac{\partial u}{\partial n}\,d\sigma(y) \\&= \int_{\partial (B_r(x) \cap B^+)} \frac{\partial u}{\partial n}\,d\sigma(y) \\&= \int_{B_r(x) \cap B^+} \Delta u(y)\,dy = 0\end{align*}

where, in the first line we used the fact that $v \equiv 0$ in $B^-$ and on the second line that $\displaystyle \frac{\partial u}{\partial n} = u_{x_n} = 0 \,$ on $\Gamma$. This completes the proof of $(3)$.